Given a real-valued function/such thatfx=tan2xx2-x2, for x>01, for x = 0x cot x , for x<0where [x] is the integral part and {x} is the fractional part of x, then
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a
limx→0+ f(x) =1
b
limx→0- f(x) =cot 1
c
cot-1limx→0- f(x) 2 = 1
d
tan-1limx→0+ f(x) π4
answer is A.
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Detailed Solution
We have limx→0+ fx= limx→0+ tan2xx2-x2 = limx→0+ tan2xx2=1 1Also, limx→0- fx limx→0- xcotx=cot1 2 x→0-, x=-1⇒x=x+1⇒x→1Also, cot-1limx→0- fx2 =cot-1cot1=1.Also, tan-1 limx→0+ fx=tan-11=π4